Derivatives of distributions in general relativity

Question

I am having some trouble when trying to reproduce some calculations involving the description of distributions (mostly used in spacetime junction conditions).

I am trying to reproduce the calculations in this paper. In the appendix, I can't derive this equality:

$$\nabla_\nu([R]n_\mu \, \underline{\delta}^\Sigma)= \underline{\delta}^\Sigma \left( [R]K^\Sigma_{\mu\nu} - [R]K^\rho_{\,\,\rho}n_\mu n_\nu +n_\mu \overline{\nabla}_\nu [R] \right) + \underline{\Delta}_{\mu\nu}, $$ where an underline designates a distribution, $\overline{\nabla}_\nu$ is the covariant derivative on the hypersurface, $[R]$ is the scalar curvature jump across the hypersurface $[R]\equiv R^+|_\Sigma- R^-|_\Sigma$, the extrinsic curvature is $$K_{\mu\nu}=h^\rho_{\,\,\mu}h^\sigma_{\,\,\nu}\nabla_\rho n_\sigma,$$ where $n$ is the normal to the hypersurface and $h$ is the projector to the hypersurface $\Sigma$, $h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu$. The distribution $\underline{\Delta}$ includes the $\delta'$ term. This is all in Eq.(37) of the paper, and the (unnumbered) equation right before it.

My problem is, instead of getting the above equality, I keep getting only $$\nabla_\nu([R]n_\mu \, \underline{\delta}^\Sigma) = \underline{\delta}^\Sigma \left( [R]K^\Sigma_{\mu\nu}+n_\mu \overline{\nabla}_\nu [R] \right) + \underline{\Delta}_{\mu\nu}. $$

If someone could point me out to what I might be doing wrong, I'd appreciate it.

The problem actually extends to eqn (38), mainly the $-2\alpha \delta$ term, which corresponds to the $f''(R)$ term in the eqn below eqn (31). I can't find where the $-K^\rho_{\,\,\rho} n_\mu n_\nu$ comes from, nor why there is no $-K^\rho_{\,\,\rho}g_{\mu \nu}$ term arising from $-g_{\alpha \beta}\nabla^\rho \nabla_\rho R$ of eqn (31).

Answer 1

The key expression that you need is eq. (163) of this paper. In that same paper, right before the equation you can find a detailed proof of it.

The equation you indicate is a direct application of it. However, although direct, there are some details to be commented, so let me go through them. By the way, let me remove the underline of the delta (since there is no tensor counterpart of it and always appears as a distribution, it can be dropped); in addition, let me call $K^\Sigma \equiv K^\Sigma{}^\rho{}_\rho $). Then: $$ \nabla_\nu([R]n_\mu \delta^\Sigma) \overset{(163)}{=} \underbrace{\nabla_\rho ([R]n_\mu n_\nu n^\rho \delta^\Sigma)}_{\underline{\Delta}_{\mu\nu}} + \Big(h^\rho{}_\nu \nabla_\rho ([R]n_\mu)- K^\Sigma n_\nu([R]n_\mu)\Big)\delta^\Sigma $$

To get your expression you have to use the chain rule in $h^\rho{}_\nu \nabla_\rho ([R]n_\mu)$. Notice that the application of the chain rule in these contexts is not trivial. For example, observe that the covariant derivative of the normal (if we see $n_\mu$ as an object defined exclusively on $\Sigma$ and not coming from a foliation) is only defined in the tangent direction. Consequently, the chain rule can be used in this case because of the $h$ that appears outside ($h^\rho{}_\nu \nabla_\rho= h^\rho{}_\nu \partial_\rho + ...$) that ensures that the partial only acts tangentially to $\Sigma$. Having said this, we apply the chain rule: $$ h^\rho{}_\nu \nabla_\rho ([R]n_\mu) = h^\rho{}_\nu \nabla_\rho [R] n_\mu + [R] h^\rho{}_\nu \nabla_\rho n_\mu $$ and finally I use in the first term that $[R]$ is a scalar, so $$ h^\rho{}_\nu \nabla_\rho [R] = h^\rho{}_\nu \partial_\rho [R] = h^\rho{}_\nu \overline{\nabla}_\rho [R] = \overline{\nabla}_\nu [R] $$ and, in the second term, the property of the extrinsic curvature $$ h^\rho{}_\nu \nabla_\rho n_\mu = K_{\nu\mu}=K_{\mu\nu}. $$

Just one subtle remark: Notice that now I have to put ${}^\Sigma$ on the extrinsic curvatures due to the fact that I have a delta in front of the expression and this evaluates the fields on the hypersurface. In addition, you have to take into account the assumption (marked with a !) that the extrinsic curvature on $\Sigma$ is chosen to be (after glueing the spacetimes) the average value taken from both sides: $K_{\mu\nu} \delta^\Sigma = K_{\mu\nu}|_{\Sigma} \delta^\Sigma \overset{!}{=} K^\Sigma{}_{\mu\nu} \delta^\Sigma$.

With all of this in mind we arrive at $$ \nabla_\nu([R]n_\mu \delta^\Sigma) = \underline{\Delta}_{\mu\nu} + \Big( [R]K^\Sigma{}_{\mu\nu}- [R] K^\Sigma n_\mu n_\nu + \overline{\nabla}_\nu [R] n_\mu \Big)\delta^\Sigma . $$